Europ.1. Combinatorics (1986) 7, 11-25

Partitions of Finite Abelian Groups

OLOF HEDEN

1. INTRODUCTION

A collection of subgroups Ot. O2, ... , On of a group 0 constitute a partition of 0 if every non-zero element of 0 is in one and only one of the groups 0 1 , O2, ... , On. We shall give conditions on the existence of partitions that consist of ni groups of order qi

i = 1, 2, ... ,k and where ql < q2 < ... < qk' We shall always assume that 0 is a finite abelian group.

If a finite abelian group 0 has a nontrivial partition then 0 is elementary abelian of type (p, p, ... , p) p a prime, see e.g. [7]. ql, q2, ... , qk and the number of elements in 0 are powers of the same prime p. If k = 2, ql = pn', q2 = pn, n> n', d = g.c.d.(n, n') and if the order of the group 0 is p2n then it is easy to prove that

n l = s(pn _1)/(pd -1), (1)

for some integer s, see Proposition 7 in Section 3. We shall show, Theorem 2 in Section 3, that if p = 2 and d = 1 then this condition also is sufficient, i.e. to each integer s with 0,;;; s,;;;(2n+ 1)/(2n'-1) there is a partition that satisfies (1). Probably, condition (1) is sufficient for every p and d. In Theorem 1 in Section 3 we shall show that for arbitrary p, nand n' (n > n') there is to every integer s with 0,;;; s,;;; m, m as in Lemma 4 in section 3, a partition which satisfies (1).

In Section 4 we shall show, using a result for mixed perfect codes, that the following condition ..

nl ~ (q2 -1)/(ql-1) (2)

is necessary. A well-known necessary condition is

nl(ql -1) + n2(q2 -1) + ... + nk(qk -1) = q -1, (3)

where q denotes the number of elements in G. We shall show in Section 5 that this condition together with the condition (2) is sufficient if 0 = GF(8) x GF(2n

), n;?; 6; OF(q) is the Galois field with q elements, and the partition contains one group of order 2n. However, there is one exception, no partition contains 5 groups of order 2, 3 groups of order 4, 2n

- 2 groups of order 8 and one group of order 2n. Other results on the existence of partitions of finite Abelian groups are to be found in

[1], [4] and [6].

2. PRELIMINARIES

If the groups 0 1 , O2, ... , On constitute a partition of 0 then each non-zero element of 0 is in precisely one of the groups 0 1 , O 2, ... , On. Hence

n

L (I Oil-1)=lol-l, i=1

where 10il denotes the number of elements of Oi' This necessary condition for existence of partitions we shall call the packing condition.

11 0195-6698/86/010011 + 15 $02.00/0 © 1986 Academic Press Inc. (London) Limited

12 0. Heden

If a finite abelian group G has a nontrivial partition then G is an elementary abelian group of type (p, p, ... , p), see [7]. One consequence of this is the following proposition which we frequently will use.

PROPOSITION 1. Suppose that G and G' are elementary abelian groups of the same type (p,p, ... ,p) and suppose that IGI=IG'I. If G has a partition containing the groups GI , G2, ... , Gn then G' has a partition containing groups G;, G~, ... , G~ where IGil = IG:I i = 1, 2, ... , n.

PROOF. According to our assumptions on G and G' there is an isomorphism qJ : G ~ G'. Put G: = qJ( G i ) for i = 1,2, ... , n. Since qJ is an isomorphism we deduce that the groups G;, G~, ... , G~ constitute a partition of G'.

The finite field with pn elements we shall denote by GF(p"). The additive group of GF(pn) is elementary abelian of type (p,p, ... ,p). By a partition ofGF(pn) we shall always mean a partition of the additive group of GF(p").

By H x K we denote the set {( h, k) I h E H, k E K}. If Hand K are additive groups we denote by {OJ x K the set {(O, k) IkE K}.

Let Hand K be two elementary abelian groups of type (p, p, ... ,p) and suppose that IKI = pn. We shall often construct partitions of H x K by using partitions of H x GF(p") and Proposition 1. If the additive groups H x {OJ and {OJ x GF(pn) are included in the partition of H x GF(p") then H x {OJ and {OJ x K are included in the partition of H x K that you get from Proposition 1.

Consider S = SI X S2 X ••. X Sn where SI, S2, ... , Sn are finite non empty sets. In each of the sets SI, S2, ... , Sn one element is denoted by O. The function

d(s, f) = l{ilsi,e ti , i = 1,2, ... , n}l,

defines a metric in S. d (S, f) is usually called the Hamming distance between sand f. A subset C of S is a perfect e-code if to every S E S there is a unique c E C with d (s, c) ,.;;; e. The elements of C are called codewords. If not all the sets SI, S2, ... , Sn have the same number of elements the code may be called a mixed perfect code. The subset

0= (0, 0, ... , 0) E s,

is called a e-sphere with centre O. If C is a perfect e-code then the minimum distance betwen codewords is 2e + 1.

PROPOSITION 2. If a finite abelian group G has a partition consisting of the groups G h G2, ... , Gn then there is a perfect I-code in G I x G2 x· .. x Gn.

PROOF. See Theorem 1 in [4].

Suppose thatthe number of elements in ni of the sets Sio S2, ... , Sn are qi, i = 1, 2, ... , m. We also suppose that ql < q2 < ... < qm and that nl + n2+· .. + nm = n. With the weight of an element s in S we mean a m-tuple (db d2, ... , dm) where di is the number of non-zero coordinates of s belonging to sets of cardinality qi.

EXAMPLE 1. Let S = SI X S2 X •.. X S5 where ISII = 4 and ISil = 2 for i,e 1. The element (ai, a2, a3 , a4 , 0) where a i ,e 0 i = 1, 2, 3,4 has the weight (3, o.

Finite abelian groups

The weight enumerator of a code is defined to be the polynomial

C(ZI>"" Zm) =L ztl ••••• Z~m,

where we sum over all the codewords c and where (dl>"" dm) is the weight of c.

13

(3')

EXAMPLE 2. Let 5 be as in Example 1. If C consists of the words (al> a2, a3, 0, 0), (0, a2, a3, a4, 0) and (0,0, a3, a4, as), where ai '" 0 i = 1, 2, ... ,5, then the weight enumerator of C is ZiZ2+2Z~.

In [3] we proved that if C is a perfect code then

m m

C(ZI,"" Zm) = Ao IT (1 + (qi -1)Zir; + L Ax IT (1 + (ql -1)Zy;-x;(1- zy; (4) I I

for some numbers Ao and Ax and where we sum over those x = (XI, , .. , xm) which satisfy a certain equation. If the perfect code is a perfect I-code then this equation is

m

1+ L [(ni -x;)(qi-1)- Xi]=O, Xi integer, O~Xi~ni' i=I,2, ... ,m. (5) i=l

Since m

151(0)1 = 1 + L ni(qi -1), i=l

equation (5) can be written

m

L xiqi = 151(0)1, Xi integer, O~ Xi ~ ni, i = 1, 2, ... , m. (5') i=l

Let P be a subset {il, ... ,ip} of{I,2, ... ,m}. To each m-tuple Y=(YI> ... ,Ym) we denote by yp the p-tuple (Yi

l,' •• , Yi)' The weight enumerator according to P of a code

we define to be

Cp(ZI,"" Zm) = C(Zil "'" Zi) = L Z~'I ... Z~p (6)

where we sum over all words in the code of weight (dl , ... , dm) where dj = 0 if j e P.

EXAMPLE 3. Let 5 and C be as in Example 2 and let P = {I}. Then

Cp(ZI, Z2) = 2zi

PROPOSITION 3. C p (1, 1, ... , 1) is the number of words in C with dj = 0 if j e P.

PROOF. If we substitute Zi. by 1, v = 1, 2, ... , p, then each term in the sum (6) will equal 1.

PROPOSITION 4. If C is a perfect I-code then the weight enumerator according to the subset P = {il> ... , pp} of {I, 2, ... , m} can be written

p P

Cp(ZI>"" Zm) = Ao IT (1 +(qi. -1)ZiJ";.+ L Ax IT (1 + (qi. -1)ZiJ";.-x'.(1-ZiJ x;. v=1 v=1

(7)

where we sum over those x = (XiI' ... , Xip

) for which x = yp for some Y that satisfies (5).

PROOF. If we substitute Zi = 0 if i e P in the equation (3) for C(ZI, ... , Zm) we get (6). If we perform the same substitution in (4) we get (7).

14 O. Heden

PROPOSITION 5. If there is a perfect I-code in S then there are at least m distinct m-tuples (XI, ... ,xm) that satisfy (5).

PROOF. See Theorem 1 in [3].

The following proposition can be proved in the same was as Theorem 1 in [3].

PROPOSITION 6. Suppose that P is a subset of {I, 2, ... , m} and that the number of elements in P is p. If there is a perfect I-code in S then there are at least p distinct p-tuples x = (Xi" . . . ,Xi.) with x ~ 0 for which i = yp for some m-tuple y satisfying (5).

EXAMPLE 4. There will not be a perfect I-code in SI x . . . X S39 where ISII = IS21 = IS31 = 2, ISil = 8 if i = 4,5, ... ,39. In this case Equation (5') will be

2Xl + 8X2 = 256, Xl and X 2 integers, 0:0.;;; Xl :0.;;; 3, 0:0.;;; x2 :o.;;; 36.

The only solution of this equation is Xl = 0, X 2 = 32. Thus by Proposition 5 or 6 and Proposition 2 GF(256) will not have a partition consisting of 3 groups of order 2 and 36 groups of order 8. This has also been proved by Bu [1]. He used a different method.

3. CONSTRUCTIONS

PROPOSITION 7. Suppose that G is a group of order pm+", m;;;. nand p a prime. If G has a partition which consists of one group of order pm, pm - X groups of order p" and y groups of order p"' where n> n' then, with d = g.c.d.(n', n),

x=s(p"'_1)/(pd_1) and y=s(p"-I)/(pd-1),

for some integers s with 0:0.;;; s:o.;;; pm(pd -1)/(p"' -1).

PROOF. From the packing condition we deduce that

x(p" -1) = y(p"' -1).

Since g.c.d.(p" -1, pn' -1) = pd -1, see, for example, the proof of 4.10 in [8], (p"­l)/(pd -1) divides y.

In this section we shall give values of s of which the necessary condition of the proposition is sufficient.

We say that a group is K'-stable if G is a subgroup of the additive group of a field K, K' a subfield of K and kg E G for each k E K' and g E G.

LEMMA 1. Let K be a finite field. Suppose that GI , G2 , ••• , Gk are K-stable subgroups of a K-stable group G and that IGil = IKI for i = 1,2, ... , Ie, Gi n q = {OJ ifi ~ j. If K' is a sub field of K and H a K' -stable subgroup of G satisfying

He. GI U·· 'uGk and IH n Gil = IK'I, i= 1,2, ... , k

then there are K' -stable subgroups H = Ho, HI,"" HI of G satisfying I Hi I = I HI i = 0,1, ... , IU:=o Hi =U;=I q and Hi n ~ ={O} ifi ~ j 1= (IKI-l)/(lK'I-1)-l.

PROOF. Let f3o, f31, ... , f31 be a system of coset representatives of the multiplicative group K'* of K' in the mUltiplicative group K* of K. We put Hi = {f3ih I h E H} i == 0,1, ... , I. Since G is K-stable the groups Ho, Hi>"" HI are in G. We first show that the union of the groups Hi i = 0, 1, ... ,I equals the union of the groups Gi i = 1, 2, ... , k.

Finite abelian groups 15

The groups Glo G2 , ••• , Gk are K-stable and thus K '-stable. Consequently this is also true for H n G; i = 1,2, ... , k. As \H n G;\ = \K'\ there is to each i i = i, 2, ... , k an a; E G; with HnG;={ka;\kEK'}. Since G; is K-stable we deduce that /3jka;EG; for each j = 0,1, ... , I and k E K'. Consider two elements x = /3,.ka; and x' = /3vk' a; where, to avoid trivialities, we assume that k and k' are non-zero elements of K'. As /30, /31, ... , /3, is a system of coset representatives of K'* in K* we conclude that if v¥- p. then x ¥- x'. Consequently

1{/3,.ka;!k E K'p. = 0,1, ... , III = (1+ l)(IK'I-l) + 1 = IKI

and thus, since \G;\ = \K\, Gj consists precisely of the elements /3jka;, k E K', and j = 0,1, .. . , I. Hence the groups G j , i = 1,2, ... , k, are subsets of the union of the groups H o, HI,"" H,. Consequently the union of the groups Glo G2 , ••• , Gk is a subset of the union of H o, HI, ... , HI. On the other hand, as the groups G I , G 2, ... , G k are K-stable, each of the groups H o, HI, ... , HI is a subset of the union of the groups G I , G2 , ••• , G k •

Hence this is also true for the union of H o, HI, ... , H,. lt remains to prove that H j n ~ = {OJ if i ¥- j. Suppose that x E H j n ~ where i ¥- j. Then

x = /3jka v = /3jk' a,. where /3j, /3j, k, k', a v and a,. are as above. We consider two cases. Case 1: v¥- p.. Then x E G v n GIL! i.e. x = O. Case 2: v = p.. Since /3j and /3j are distinct coset representatives we deduce that k = k' = O. Consequently x = O.

Let G = K x K where K is a finite field. lt is easy to check that the sets K", = {(k, ak)\kE K}, a E K, are subgroups of K x K which together with the group {O}x K constitute a partition of K x K. We denote this partition by '/TK'

Suppose that K' is a subfield of K and that K is a finite field. The group of automorph­isms of Kover K' is cyclic, see [5, p. 185] Let u denote an element that generates this group. To each K'-stable subgroup H of K we define a K'-stable subgroup Her of G=KxK by

LEMMA 2. If G' E IlK then either G' n Her = {OJ or \G' n H U\ = \K'\.

PROOF. Supppose that (x, u(x» and (y, u(y» with x¥-O and y¥-O are in the same group K", of the partition IlK of G. Then

u(x) u(y) --=--=a.

x y

Hence u(x)1 u(y) = xl y, i.e. u(xl y) = xl y. Since u generate the Galois group of Kover K' we deduce that xl y belongs to the fixed field of the Galois group, i.e. xl y E K'. Then x = yk for some k E K'. The number of elements in K", n H U is consequently at most \K'\.

If (x, u(x» E K", then, since u(kx) = u(k)u(x) = ku(x), (kx, u(kx» E K", for each kE K'. Consequently the number of elements in K", n H eT is 1 or at least IK'I·

LEMMA 3. If H l nH2={0}, HI and H2 K'-stable, K",EIlK and \HfnK",\> 1 then

\Hrn K",\ = 1.

PROOF. According to the assumption there is xEHI\{O} with u(x)/x=a. If \Hin K", \ > 1 then there is y E H2 \ {OJ with u(y)1 y = a. As in the proof of Lemma 2 we deduce that x = ky for some k E K'. As H2 is K' -stable we conclude that x E H2 which gives a contradiction.

Since the additive group of every field K has K' -stable subgroups, K' a subfield of K, it is possible, by using Lemma 1 and 2, to construct partitions of K x K which consists

16 O. Heden

of groups of two or more distinct orders. If we use Lemma 3 and the following lemmas we get a better result.

LEMMA 4. Suppose that d divides both nand n', n > n', and that n = kn' + r where n',,;;: r< 2n'. Let

_ {pn-n' + pn-2n' + ... + pn-kn' + 1, m - (pn -l)/(pn' -1),

if r ¥- n',

if r = n',

Then GF(pn) has a partition II which consists of one group of order pr and m -1 GF(pd)_ stable groups HI, H 2, ... , Hm+l each of order pn'.

PROOF. Case 1: r= n'. Then n' divides n and hence GF(pn') is a subfield ofGF(pn). In the same way as in the proof of Lemma 2 in [4] we construct a partition of GF(pn). pn' is the order of the groups in this partition. From the construction in [4] it follows that these groups are GF(pn')-stable and thereby also GF(pd)-stable. Case 2: r¥- n'. Consider the additive group of GF(pn) as a vectorspace over GF(pd). According to the paragraph after the proof of Lemma 4 in [1], this vector space has a partition II which consists of one subvectorspace of dimension rl dover GF(pd) and m -1 subvectorspaces of dimension n'l dover GF(pd). These vectorspaces are GF(pd)-stable subgroups of GF(pn) and contain pr respectively pn' elements.

LEMMA 4'. With the same assumptions as in Lemma 4, GF(p") has a partition II' that consists of m GF(pd)-stable groups HI> H 2, ... , Hm each of order pn' and groups of order pd.

PROOF. If r = n' this is trivial by Lemma 4. If r ¥- n' then r> n'. The vector space of dimension rl dover GF(pd) in the partition II of the proof of Lemma 4 has a partition in one GF(pd)-stable group Hm of order pn' and groups of order pd.

THEOREM 1. Suppose that n = kn' + r where n' < r < 2n' and suppose that d divides both nand n'. Let m be as in Lemma 4. To each integer s with 0,,;;: s";;: m GF( pn) x GF(pn) has a partition that besides {O}xGF(pn) and GF(pn)x{O} consists of pn_1_ s(pn' _1)/(pd -1) groups of order pn and s(p" _1)/(pd -1) groups of order pn'.

PROOF. By Lemma 4', GF(pn) has m GF(pd)-stable subgroups HI> H 2, ... , Hm with H j II Hj = {O} if i ¥- j. Let IlK be defined as in the paragraph before Lemma 2. By Lemma 3, the groups HI> H 2 , ••• , Hm divides the groups of IlK in m distinct subclasses SI> S2,"" Sm which by Lemma 2 satisfies

The number of groups in a class Sj is (pn' _1)/(pd -1). The union of these groups can, according to Lemma 1, be substituted by (pn _1)/(pd -1) groups of order pn' which in pair only has the zero in common. "If we carry out this substitute in s of the classes SI, S2, ... , Sm we get the theorem.

REMARK. If r = 0 then n' divides n. The situation is then almost trivial. In the partition IlK of K x K, where K = GF(pn), each group has a partition in (pn _1)/(pn' -1) groups of order pn'.

Examples indicate that the theorem is true for each integer s with 0,,;;: s ,,;;: p"(pd -1)/(pn' -1). We show that this is the case if p = 2 and d = 1.

Finite abelian groups 17

THEOREM 2. Suppose that n=kn'+r where O<r<n'. GF(2n)xGF(2n) has for each integer s where 0 ~ s ~ (2n - 2r )/(2n' -1) a partition that besides {O} x GF(2n) and GF(2n) = {O} consists of s(2n -1) groups of order 2n' and 2n -1- s(2n' -1) groups of order 2n.

PROOF. Let h be any element of GF(2n) and let (Th denote the map

(Th :x~ hx+x2

from GF(2n) to GF(2n). As (Th is a homomorphism we conclude that for any subgroup H of the additive group of GF(2n) the subset

HO"h = {(x, O"h(X)) 1 x E H}

of GF(2n) x GF(2n) is a group. Consider the partition IlK of GF(2n) x GF(2n), defined in the paragraph following the proof of Lemma 1 in this section, and the subgroups K a ,

a E GF(2n), of this partition. An element (x, (Th(X)), x,e 0, of HO"h belongs to Ka if and only if (Th(X) = ax, i.e. from the definition of (Th, a = h + x. Further

We conclude that

and that

HO"h (\ ({O} xGF(2n)) = {(O, On.

for aEh+(H\{O})

HO"hC U Ka. aEh+(H\{O})

(If h happens to belong to H\{O} then IHO"h (\ Kol = 2. Below we shall always choose h such that h does not belong to H\{O}.)

Now we can use Lemma 1 to substitute all the groups Ka where aEh+(H\{O}) by subgroups with the same order as H and that only have the element (0,0) in common. If we choose subgroups H of GF(2n) and elements h of GF(2n) in such a manner that the sets h + (H\{O}) mutually are disjoint we can perform many such substitutions.

Let G denote the additive group of GF(2n). As n = kn ' + r there are subgroups H}, H2 , ••• , Hk of G, each of order 2n', and a subgroup H' of G of order 2r such that

G = H) X H2 X ••• X Hk X H'.

As distinct co sets of H) are disjoint we conclude that the sets

h+(H)\{O})h E {O} x H 2 x· .. X Hk X H'

are mutually disjoint. Further all elements of G except the elements of the group {O} x H2 x ... X Hk X H' belong to the union of these sets. Proceeding in the same way we consider for i E {1, 2, ... , k} the subsets

h+(Hi\{O}) where hE{O}x·· ·x{0}XHH1 •• ·xHkxH', i~k

of G. These sets are mutually disjoint and the number of such sets equals (2n - 2r)/ (2n' -1).

Hence the theorem follows from Lemma 1. We now consider partitions of GF(pn) x GF(pm), where m> n, that beside the groups

{O} x GF(pm) and GF(pn) x {O} contain groups of order pn respectively pn', n> n'. The following lemma will be of great importance in the sequel.

LEMMA 5. Suppose that G = H x K and that the group K has a partition which consists of the groups K}, K 2 , ••• , Kn. If each of the groups H x Ki, i = 1, 2, ... , n, has a partition II; that besides H x {O} and {O} x Ki consists of groups L:,j = 1,2, ...• mi. then

18 0. Heden

the groups H x {OJ, {OJ x K and the groups LJ, i = 1,2, ... , n, j = 1,2, ... , mj constitute a partition II of G.

PROOF. We show that G is the union of the groups in II. That the intersection of two distinct groups only is the element (0,0) can be proved in a similar manner.

The elements (h,O) and (0, k) hE H, k E K are in the groups H x {OJ and {OJ x K. If h .,t. 0 and k.,t. 0 then (h, k) belongs to one of the groups H x K j and thus to one of the groups LJ.

THEOREM 3. Suppose that t = kn + r where k;;:. 1 and n < r < 2n and suppose that n = k'n' + r' where n' < r' < 2n'. Let d = g.c.d.(n, n') and put

m' = pn-n' + pn-2n' + ... + pn-kn' + 1

m = p,-n + p'-2n + ... + p'-kn + 1.

The group GF(pn) x GF(p') has for each integer s with 0,,;;; s";;; m· m' a partition that besides {OJ x GF(p') and GF(pn) x {OJ contains p' -1- s(pn' _1)/(pd -1) groups of order pn and s(pn _1)/(pd -1) groups of order pn'.

PROOF. According to Lemma 4', GF(p') has a partition that consists of m groups K}, K 2, ... , Km of order pn and one group K of order pro Each group GF(pn) x K j ,

i = 1, 2, ... ,m, has, by Theorem 1 and Proposition 1 in Section 2, for every s' with 0,,;;; s',,;;; m' a partition that besides {OJ x K j and GF(pn) x {OJ consists of s'(pn _1)/(pd -1) groups of order pn' and pn -1- s'(pn' _1)/(pd -1) groups of order pn. The group GF(pn) x K has a partition that consists of {OJ x K, GF(pn) x {OJ and groups of order pn. (We construct this partition in the same way as the partition IlK is constructed. Let H be a subgroup of order pn of GF(pr). Put Ha = {(h, ah) I h E H} a E GF(pr). {OJ X GF(pr) and the groups Haa E GF(pr) constitute a portion of H x GF(pr). Since IHI = IGF(pn)1 and IGF(pr)1 = IKI we can use Proposition 1 in Section 2.)

By Lemma 5 the theorem is proved. If n divides t we can do even more.

THEOREM 3'. Suppose that t = kn where k;;:. 2 and suppose that n = k' n' + r' where n' < r' < 2n'. Let d = g.c.d.(n, n') and let m' be as in Theorem 3. Put

m = (p' _1)/(pn -1).

The group GF(pn) x GF(p') has for each integer s with 0,,;;; s,,;;; m· m' a partition that besides {OJ x GF(p') and GF(pn) x {OJ contains p' -1- s(pn' _1)/(pd -1) groups of order pn and s(pn _1)/(pd -1) groups of order pn'.

The proof is the same as the proof of Theorem 3. In the two theorems above we considered partitions of GF(pn) x GF(p') where t;;:. 2n.

The next lemma can be used to construct partitions where n < t < 2n.

LEMMA 6. Suppose that G = G F( P n) X G F( p') where 2 < n ,,;;; t and suppose that G has a partition II that besides {OJ x GF(p') and GF(pn) x {OJ consists of groups of order pn and groups of order pn-l. Let G' = H X GF(p') where H is a subgroup of GF(pn) of order pn-l.

The groups G'II L L E II constitute a partition II' of G'. II' contains besides H x {OJ and {OJ x GF(p') groups of order pn-l and of order pn-2. the number of groups in II' of order pn-2 does not depend on the choice of the group H'.

The construction in this lemma is the same as in Lemma 5 of [1]. The proof also will almost be the same.

Finite abelian groups 19

PROOF. Trivially, O'n(GF(p")x{O})=Hx{O} and O'n({O}xGF(p'»= {O} x GF( p').

Note that if LE II\( GF( p") x {O} u {O} x GF( pi» then those elements h of GF( p") for whicn there is kh E GF(p') with (h, kh) E L consitute a subgroup of GF(p") with the same order as L, else L should contain an element (0, k) where k ¥- 0, which contradicts our assumption on L. We denote this subgroup of GF(p") by LI.

GF(p") is a vectorspace over GF(p) of dimension n. If H is a subspace of GF(p") of dimension n -1 and L' another subspace of GF(p") then

dim L' n H = dim L' or dim L' - 1.

By this and the observation above we conclude that II' besides H x {O} and {O} x GF( p') consists of groups of order p"-I or p"-2.

By the packing condition the number of groups in II is uniquely determined by the number of groups in II of order p"-I. On the other hand, the number of groups of II' of order p"-2 is uniquely determined by the number of groups of II'. Since n > 2, II and II' contains the same number of groups. Now the proof of Lemma 6 is complete.

REMARK. If n = 2 then the situation is trivial. Every abelian group of order p2 and type (p, p) has a partition that consists of p + 1 groups of order p.

From Lemma 6 we can deduce results analogous to those in Theorem 1,2 and 3. We give an example that we need in Section 5.

EXAMPLE. Let 0 = GF(16) x GF(16). By Theorem 2, G has one partition III which consists of GF(16) x {O}, {O} x GF(16), 8 groups of order 16 and 15 groups of order 8. 0 has also one partition II2 which consists of GF(16) x {O}, {O} x GF(16), one group of order 16 and 30 groups of order 8.

Let H be a subgroup of GF(16) of order 8 and put 0' = H x GF(16). The groups 0' n L, L E II; i = 1 or 2 constitutes a partition II; of 0'. II; contains besides H x {O} and {O} x GF(16): 9 groups of order 8 and 14 groups of order 4, if i = 1; 3 groups of order 8 and 28 groups of order 4, if j = 2.

4. SOME NECESSARY CONDITIONS

The set of groups of least order in a partition II of a group 0 we shall call the tail of the partition. The number of groups in the tail we shall call the length of the tail. In this section we show that the tail of a partition has a certain minimal length.

Suppose that the partition II of 0 consists of n; groups of order qi, i = 1,2, ... , m, and suppose that ql < q2 < ... < qm·

LEMMA 7. If a finite abelian group 0 has a partition II then

nl ;;' q2/ ql·

PROOF. There is a perfect I-code in the direct product of the groups in the partition II, Proposition 2 in Section 2. According to Proposition 6 in the same section there is at least one m-tuple (XI, X 2 , ••• , xm) with XI ¥- 0 and 0 ~ Xi ~ ni , i = 1,2, ... , m, satisfying

m

Xl ql+X2q2+ I xmqm=ISI(o)l· (8)

Since 0 is a finite abelian group, q; = p" for i = 1,2, ... , m, where p is a prime.ISI(O)1 = p' since ISI(O)I = 101. Since tl < t2 < ... < tm < t we deduce from Equation (8) that q2 divides

20 0. Heden

xlql· Consequently q2/ ql divides XI. Since XI ¢ 0 we must have xl;3 q2/ ql and thus nl ;3 q2/ ql·

THEOREM 4. If a finite abelian group has a partition II then

PROOF. Suppose that G has a partition II with nl < (q2 - I)/(ql - 1). In the direct product of the groups in the par.t~tion there is a perfect I-code, Proposition 2 in Section 2. If an m-tuple (XI, X 2, ••• , xm) satisfies Equation (5') in Section 2 then, by the proof of Lemma 7 q2/ ql divides XI. Further, from our assumption on nl and by Lemma 7, we deduce that q2/ ql"; nl < 2q2/ ql. Now, by Proposition 4 in Section 2, the weight enumerator according to the subset {I} of {I, 2, ... , m} may be written

(9)

where XI = q2/ ql and where A and B are constants. As the minimum distance between codewords is 3 and as the word (0, 0, ... ,0) belongs

to the code there are no codewords of weight (dl , 0, ... ,0) where dl = 1 or 2. Thus

C(Zd = 1 + terms of degree;3 3. (10)

If we calculate the coefficients of 1 and ZI in (9) and use (10) we get two equations for A and B. These equations have the solution

(11)

By Proposition 3 in Section 2 we now get that the number of words of weight (dl , 0, ... ,0) in the code is

(12)

On the other hand, since the minimum distance between codewords' is 3, we can have at most q~l/(1 + nl(ql -1)) such words. If n l < (q2 -1)/(ql-l) this number if according to (12) less than C(l), which gives a contradiction.

The inequality in the theorem may not be improved generally. GF(p4), i.e. has a partition that consists of p2+ 1 groups H;, i = 1,2, . .. , p2+ 1, each of order p2. HI has a partition consisting of p + 1 groups L; i = 1, 2, . .. , p + 1 each of order p. The groups LI> ... , Lp+1> H 2, .. . , H p2+J constitute a partition with nl = (q2 -1)/ (ql -l).

In [6] Lindstrom proved that if a partition of a finite abelian group consists of one group of order pa and groups of order pb then a;3 b. This result follows from Theorem 1 since we never have nl = 1.

We need the following result in Section 5.

THEOREM 5. A finite abelian group G never has a partition II with n; groups of order 2;, i = 1, 2, ... , m, where n l = 5 and n2 = 3.

PROOF. Suppose that G has such a partition. By Proposition 2 in Section 2, there is a perfect I-code in the direct product of the groups in the partition. By Proposition 5 in Section 2 there is at least m m-tuples satisfying

m

2XI+ 4x2+ L ix;=2', (13) i=3

where ISI(O)I = 2'. From this equation we deduce that XI = 0 (mod 2) and if XI = 0 then X2 = 0 (mod 2). Further if (XI> X2, •.• ,xm ) and (x;, x~, ... , x:..) are two m-tuples satisfying

Finite abelian groups 21

(13) and Xl = X; then X2 == x~ (mod 2). Now the weight enumerator according to the subset {I,2} of {I, 2, ... , m} can, by Proposition 4 in Section 2, be written

C(Z, U) = A(l + 3Z)3(l + U)5 + A'(l + 3Z)(I- Z)2(1 + U)5

+ B(1 + 3Z2)( 1- Z)( 1 + U)3(1- U? + B'(1- Z)3(1 + U)3(1- U)2

+ C(1 + 3Z)3(I + U)(I - U)4+ C(I + 3Z)(I- Z)2(1 + U)(1- U)4 (14)

(Z] = U and Z2 = Z) for some constants A, A', B, B', C and C. As in the proof of Theorem 4 we get that

C(Z, U) = 1 + terms of degree;;;. 3.

Thus the coefficients of Z, U, Z2, U 2 and ZU in (14) are zero. If we use this we can calculate the constants A, A', ... , C'. We get that A = 1/64, A' = 3/64, B = 12/64, B' = 28/64, C=-I/64 and C=2I/64. If we use these values of A,A', ... ,C' and calculate the coefficient of Z2 U 2 in (14) we get -6 which is impossible.

EXAMPLE. Consider GF(64) and a partition of GF(64) that consists of groups of order 2, 4 respectively 8. Let n] , n2 and n3 be as above. Theorem 4 excludes the following 6 possibilities for (nlo n2, n3)

(1,2,8), (2,4,7), (1, 9, 5), (2,11,4), (1,16,2) and (2,18,1).

These possibilities are, however, not excluded by the packing condition. Theorem 5, but not Theorem 4 or packing condition, exclude (nlo n2, n3) = (5, 3, 7). For all the remaining 3-tuples (nlo n2 , n3 ) that satisfies the packing condition there is a partition as we shall see in the next section.

5. PARTITIONS OF GF(8) x GF(2") WHERE n > 2

THEOREM 6. GF(8) x GF(2") n;;;. 6 has a partition that besides GF(8) x {OJ and {OJ x GF(2") consists of ni groups of order 2i i = 1,2 or 3 if and only if nlo n2 and n3 satisfy the following two conditions:

(a) nl+3n2+7n3=7· (2"-1);

(b) if n2 r= 3 then nl = 0 or n];;;' 3; if n2 = 3 then nl > 5.

That the condition (a) is necessary follows from the packing condition, since the number of elements in the set GF(8) xGF(2")\(GF(8) x{0}u{0}xGF(2"» is 7· (2"-1). From the results of Section 4, as in the example of that section, it follows that condition (b) is necessary. It remains to prove that the conditions (a) and (b) are sufficient.

Let n], n2 and n3 be as in Theorem 6. We shall say that a partition II of H x K where IHI = 8 and IK 1= 2" n;;;' 3 is of type PI if

n3 =2" -I-t,

{7t/3,

n2 = 7 ( t - 1) /3 + 1,

n2 = g(t- 2)/3+3

if t==0(mod3),

if t== 1 (mod 3),

if t == 2 (mod 3), t r= 2,

if t == 2.

LEMMA 8. IfG == GF(8) x GF(2") n;;;. 6 has a partition of type Prfor t = 0,1, ... ,2"-1 then for each 3-tuple (n l , nz, n3) that satisfies condition (a) and (b) of Theorem 6 there

is a partition of G that besides GF(8) x {OJ and {OJ x GF(2") consists of ni groups of order 2i, i == 1, 2, or 3.

22 0. Heden

PROOF. Let II and II' be partitions of G that contain the same number of groups of order S. Suppose that the number of groups in II of order 2i, i = 1 or 2, is ni and that the corresponding numbers of II' are ni, i = 1 or 2, and suppose that n; < n2. Since II and II' contains the same number of groups of order S

nl+3n2=n~+3n;.

We deduce that 3 divides n~ - n1 • From II we can construct a partition with n~ groups of order 4 by splitting up (n~ - n1)/3 of the groups of order 4 in groups of order 2.

The partitions of type P, are optimal in the sense that there do not exist partitions which for a given number of groups of order S contains more groups of order 4 than those of type Pro This follows from the results of Section 4.

To show that conditions (a) and (b) of Theorem 1 are sufficient it is, according to Lemma S, enough to prove the following theorem.

THEOREM 7. GF(S) x GF(2") n ~ 6 has a partition of type P, for t = 0,1, ... , 2"-1.

We shall prove this theorem by induction. However, we need a couple of lemmas.

LEMMA 9. IfGF(S) x GF(2"), n ~ 3, has a partition of type PJor t == 0 (mod 3) t ~ 2"-2 then G has a partition of type P,+ l' G always has partitions of types Po, P1 and P2.

PROOF. Let II be a partition of type P, t == 0 (mod 3). If H is a group of order S in II then H has a partition consisting of one group of order 4 and 4 groups of order 2. If we in II substitute H by these groups we get a partition of type P,+l'

In the same way we deduce that Gf(S) x GF(2") always has partitions of type Po, P1

and P2 •

LEMMA 10. G = GF(S) x GF(S) has a partition of type PJor t = 0,1,2, ... ,7.

PROOF. From Theorem 2 in Section 3 and Lemma 9 above we deduce that G has a partition of type P, for t = 0, 1, 2, 3, 4, 6 and 7. We now construct a partition of type Ps.

Let e denote an element of GF(S) that satisfies e3 + e + 1 = O. Then e is a primitive element of GF(S), i.e. GF(S) = {O, eO, e\ ... , e6}. Put

i = 5 or 6.

Hs, H 6 , GF(S) x {O} and {O} x GF(S) constitute together with the following groups a partition of type Ps . In the enumeration below we write i instead of e i; only elements which are not equal to (0,0) will be indicated:

{(O, 2), (1,3), (3,5)}, {(O, 4), (1,5), (3, O)}, {(I, 2), (2,4), (4, I)}

{(2, 2), (3,4), (5, I)}, {(3, 3), (4,5), (6, 2)}, {(4, 4), (5,6), (0, 3)}

{(4, 6), (5,2), (O,O)}, {(5, 5), (6,0), (I, 4)}, {(5, 0), (6,3), (I, I)}

{(6, 6), (0, I), (2, 5)}, {(3, 6)}, {(4, O)}, {(6, I)}, {(2, 6)}, {(2, 3)}.

LEMMA 11. G = GF(S) x GF(16) has a partition of type P, for t = 0, 1, 2, 4, 5, 6, 7, S, 11, 12, 13, 14 and 15.

REMARKS. It is not known to the author of this paper if G has partitions of type P3 ,

P9 and PlO • Probably there is a partition of type PlO and possibly also of type P9 •

Finite abelian groups 23

PROOF. From the example of Section 3 and from Lemma 9 it follows that G has partitions of type P, for t = 1, 2, 6, 7, 12 and 13. GF(16) has a partition that consists of 5 groups of order 4. If we use this partition and apply Lemma 5 in Section 3 we get a partition of type P15 •

Let e be an element ofGF(16) that satisfies e4 + e + 1 = O. From [2] we deduce that the set H = {O, eO, e\ e2

, e\ e5, elO, eS

} is a subgroup of GF(16). We shall now construct partitions of H x GF(16). As Hand GF(8) have the same number of elements we get, according to Proposition 1 in Section 2, from each partition of H x GF(16) a partition of GF(8) x GF(16). Put Hi = {(h, eih) I h E H}, i = 0,1, ... ,14. Consider the following 10 subsets of H x GF(16) (we write i instead of e i

);

MI = {(O, 9), (1, 10), (4, 13)}, M2 = {(8, 2), (0,11), (2, 9)},

M3 = {(1, 12), (2, 13), (5, I)}, M4 = {(I, 7), (2,11), (5, 8)},

Ms = {(2, 5), (4,7), (10, 13)}, M6 = {(2, 8), (4,0), (10, 2)},

M7 = {('I, 10), (5, 14), (8, IT)}, Ma = {(5, 12), (10,4), (0,6)},

M9 = {(IO, 6), (8, 14), (1, 8)}, MIO = {(10, I), (8,0), (1, 4)}.

Gi = Mi U {(O, O)} is a subgroup of H x GF(16), i = 1, 2, ... ,10, and Gi C H3 U H6 U H7 U

H9 U H 11 , for i = 1, 2, ... , 10. Further Gi n Gj = {(O, O)} if i,e j. Every element in the set (H3 U H6 U H7 U H9 U H11)\U:~1 Gi constitute together with (0,0) a subgroup of H x GF(16). These groups of order 2, the groups G 1 , G2 , ••• , G IO , the groups Hi for i = 0, 1,2,4,5,8,10,12,13 and 14, H x {O} and {O} x GF(16) constitute a partition of H x GF(16). This partition is of type P5 •

We get a partition of type Ps in a similar way using the following 17 subsets of HxGF(16).

NI = {(8, 5), (0, 12), (2, 14)},

N3 = {(5, I), (2,8), (1, 10)},

Ns = {(5, 14), (2,1), (I,7)},

N7 = {(I, 12), (8,2), (10, 7)},

N9 = {('I, 3), (1,4), (0,7)},

Nl1 = {(10, 4), (4, IT), (2, 13)},

N\3 = {(O, 6), (10,2), (5,3)},

N 1s = {(O, 13), (10, I), (5, 12)},

N17 = {(8, 0), (5,8), (4,2)}.

N2 = {(5, 4), (2,5), (1, 8)},

N4 = {(I, 0), (8,7), (10, 9)},

No = {('I, 0), (1, 14), (0,3)},

Na = {(10, 6), (4, 13), (2, O)},

NIO = {(O, 14), (10, 13), (5, 2)},

N12 = {('I, 10), (I, 13), (0,9)},

NI4 = {(10, 8), (4,7), (2, IT)},

N I6 = {(8, 6), (5, IT), (4,1)},

All these sets are namely subsets of 8 of the groups Ho, HI, ... , H 14 •

It remains to construct partitions of type P4 , P11 and P14 • Partitions of type P11 and P14 we get by completing Mt. M 2 , ••• , M IO , and Nt. N 2 , ••• , N 17 , respectively, with the following 14 sets.

LI = {(O, 0), (1,2), (4,8)},

L3 = {(I, 1), (2, 12), (5, 13)},

Ls = {(2, 3), (4, 12), (10, 10)},

L7 = {('I, 4), (5,5), (8,8)},

L9 = {(5, 6), (10, IT), (0, I)},

Ll1 = {(10, 14), (8, 12), (I, 5)},

L\3 = {(8, 9), (0,5), (2,6)},

L2 = {(O, 8), (1,6), (4, 14)},

L4 = {(I, 11), (2,2), (5, 9)},

L6 = {(2, 10), (4,5), (10, O)},

La = {('I, 9), (5, 10), (8, TI)},

LIO = {(5, 0), (10, 5), (0, 10)},

L12 = {(10, 3), (8,1), (I,9)},

LI4 = {(8, 3), (0,4), (2,7)}.

24 o. Heden

We leave it to the reader to construct a partition of type P4 • We shall not need such a partition in the future.

In the proofs below we shall use Lemma 5 in Section 3. To avoid repetition we shall not refer to this lemma every time it will be used.

Note that if we choose partitions of type P" in GF(8) x K j , i = 1, 2, ... , n, and apply Lemma 5 in Section 3 we will not always get a partition of type P ',+'2+ ... +'". The partition will be of type P ,,+'2+ ... +'" if and only if at least n -1 of the numbers tl , t2 , ••• , tn are divisible by 3.

If IKI = 4 and IHI = 8 then H x K has a partition which besides H x {O} and {O} x K consists of 7 groups of order 4. We shall say that this partition is of type L.

LEMMA 12. 0 = GF(8) x GF(64) has a partition of type P,for t = 0,1,2, ... ,63.

PROOF. GF(64) has a partition that consists of9 groups Kt. K2 , ••• , K9 each of order 8. By Lemma 10, GF(8) x K j has partitions of type Po, P3 , Ps and P6 • If we choose partitions of type Po, P3 or P6 in GF(8) x K j for i = 1,2, ... , 9 we get partitions of type P, in 0 where t == 0 (mod 3) and t",; 54. If we choose a partition of type Ps in GF(8) x KI and partitions of type Po, P3 and P6 in GF(8) x K j for i = 2, 3, ... ,9 we get partitions of type P, where t == 2 (mod 3) and 5",; t",; 53. If we apply Lemma 9 we get partitions of type P, for the remaining values of t where 0",; t",; 55.

GF(64) has a partition that consists of one group H of order 16 and 16 groups K I, K2 , ••• ,K16 each of order 4. We choose a partition of type L in GF(8) x K j ,

i = 1, 2, ... , 16, and partitions of type P, where t = 8, 11, 12, 13, 14, 15 in GF(8) x H. From these partitions we get partitions of types P, of 0 where t = 56, 59, 60, 61, 62, 63.

From Lemma 9 we deduce that all that remains to be done is to construct a partition of type PS7 . From Theorem 2 in Section 3 we know that G F( 64) has a partition consisting of 3 groups each of order 8 and 14 groups L I, L2 , ••• , LI4 each of order 4. In GF(8) x Li, i = 1, 2, ... , 14, we choose partitions of type L. In GF(8) x K j , i = 1, 2, ... , 14, we choose partitions of type L. In GF(8) x K j , i = 1 and 2 we choose partitions of type P6 and in GF(8) x K3 a partition of type P3. Applying Lemma 5 we get a partition of type PS7 .

LEMMA 13. 0 = GF(8) x GF(128) has partitions of type P, for t = 0, 1,2, ... , 127.

PROOF. GF(128) has a partition that consists of one group K of order 16 and 16 groups K I, K2 , ••• , KI6 of order 8. If we choose a partition of type Po in GF(8) x K and suitable partitions of type P, in GF(8) x K j , i = f, 2, ... , 16, we get, as in the proof of Lemma 12 partitions of type P, of 0 for 0",; t",; 97. If we choose a partition of type PIS in GF(8) x K and partitions of GF(8) x Ki, i = 1,2, ... ,16, as above we get partitions of type P, of 0 for t = 15, 16, 18, 19, 20, 21, ... , 112.

For the remaining values of t, we choose a partition of GF(128) that consists of one group K of order 16, one group K' of order 8 and 35 groups Lt. L2 , ••• , L3S each of order 4. This is a partition of type PIS of 0' x 0" where 10'1 = 8 and 10"1 = 16. We choose partitions of type L in the groups GF(8) x L j , i = 1, 2, ... ,35, and partitions of GF(8) x K and GF(8) x K' according to the following table:

Partition of GF(8) x K of type P, where t = 5 6 6 6 6 7 8 12 12 12 12 13 14 15 15

Partition of GF(8) x K' of type P, where t = 3 3 4 5 6 6 6 3 4 5 6 6 6 6 7

We then get partitions of type P, where t = 113, 114, ... ,127.

Finite abelian groups 25

LEMMA 14. Suppose that n = n'+2, n';?;6. If GF(S) xGF(2",) has partitions of type p. for t = 0,1,2, ... ,2"' -1 then 0= GF(S) x GF(2") has partitions of type p. for t = 0,1,2, ... ,2"-1.

PROOF. GF(2n) has a partition that consists of one group K of order 2n-

3 and 2"-3 groups K;, i = 1, 2, ... ,2"-3, of order S. GF(S) x K has a partition of type Po. Ifwe choose partitions of suitable types P, in GF(S) x K;, i = 1, 2, ... ,2"-\ as in the proof of Lemma 12, we get partitions of type p. for t = 0, 1,2, ... , 6 . 2"-3 + 1.

GF(2") has a partition that consists of one group K' of order 2"-2 and 2n-

2 groups L; i = 1,2, ... ,2"-2 of order 4. In the groups GF(S) xL;, i = 1, 2, ... ,2"-2, we choose parti­tions of type L. According to our assumptions, GF(S) x K' has partitions of type P, for t = 0, 1,2, ... , 2"-2 -1. By using Lemma 5 of Section 3, we get partitions of a of type p. for t = 3· 2"-2,3' 2"-2+ 1,3' 2"-2+ 3,3' 2"-2+4, ... ,2"-1.

Finally, we have to construct a partition of type P, where t == 6 . 2"-3 + 2. Let K and K;, i = 1, 2, ... ,2"-3, be as above. As n;?; S K has a partition which consists of one group K" of order 2"-5 and 2"-5 groups L;, i = 1, 2, ... ,2"-5, each of order 4. If w.e choose a partition of type Po in GF(S) x K" and partitions oftype Lin GF(8) xL;, i == 1,2, ... ,2"-5, we get a partition of type P" t = 3 ·2"-5 in GF(8) x K. Using this partition and suitable partitions of type p. of the groups GF(8) x K;, i = 1, 2, ... , 2"-3, we get partitions of type P, for 3 . 2"-5 + 3:0:;; t ~ 3 . 2"-5 + 6 . 2"-3 + 1. 6 . 2"-3 + 2 is surely in this interval.

PROOF OF THEOREM 7. According to Lemmas 12 and 13, the theorem is true if n = 6 and n = 7. By Lemma 14 and induction, the theorem is true for every n;?; 6.

REMARK. By lemma 2 in [6] if a finite abelian group a has a partition that consists of the groups 01> O2 , ••• , a" then lad· 1 ql:o:;; 101 for every i and j, i ¥- j. Consequently if 0= GF(8) x GF(2") has a partition II that contains one group of order 2" then the remaining groups in the partition have orders 2, 4 or 8. Thus Theorem 6 gives a necessary and sufficient condition for existence of partitions of a that contains n 1 groups of order 2, n2 groups of order 4, n3 + 1 groups of order 8 and one group of order 2".

REFERENCES

1. T Bu, Partitions of a vector space, Discrete Math. 31 (1980),79-83. 2. W. H. Bussey, Galois field tables for p".;;; 169, Bull. Amer. Math. 12 (1905,,1906), 22-38. 3. O. Heden, A generalised Lloyd theorem and mixed perfect codes, Math. Scand. 37 (1975), 13-26. 4. M. Herzog and J. Schonheim, Group partition, factorization and the vector covering problem, Canad. Math.

Bull. 15(2) (1972),207-214. 5. S. Lang, Algebra, Addison-Wesley, 1971. 6. B. Lindstrom, Group partitions and mixed perfect codes, Canad. Math. Bull. 18(1) (1975), 57-60. 7. G. A. Miller, Groups in which all operators are contained in a series of subgroups such that any two have

only identity in common, Bull. Amer. Math. 12 (1905-1906),446-449. 8. R-H. Schulz, Uber Blockpliine mit transitiver Dilatationsgruppe, Math. Zeitschr. 98 (1967), 60-82.

Received 22 March 1982 and in revised form 18 June 1984

OLOF HEDEN

Department of Mathematics Royal Institute of Technology,

S-10044 Stockholm, Sweden